JEE Main · Mathematics → Steady

Statistics appeared 21 times across 3 years — 2.4% of Mathematics. This question is from Mean and Variance.

Year 2026 2025 2024 Total
Questions 7 7 7 21

If the mean and the variance of the data
Class4–88–1212–1616–20
Frequency3λ47
are μ and 19 respectively, then the value of λ + μ is

Solution & Explanation

Related Formula
μ = (Σ fᵢ xᵢ)/(Σ fᵢ) σ² = (Σ fᵢ xᵢ²)/(Σ fᵢ) - μ²
Core Logic

Midpoints (xᵢ): 6, 10, 14, 18. Frequencies (fᵢ): 3, λ, 4, 7. Sum of frequencies Σ fᵢ = 3 + λ + 4 + 7 = 14 + λ.

Calculate Mean (μ):

μ = (6(3) + 10(λ) + 14(4) + 18(7))/(14 + λ) μ = (18 + 10λ + 56 + 126)/(14 + λ) = (10λ + 200)/(λ + 14) = 10 + (60)/(λ + 14)

Since data is standard, μ and λ typically hold integer values, hinting λ + 14 must be a factor of 60. Therefore λ could be 1, 6, 16.

Step 1: Applying Variance

Variance σ² = 19.

σ² = (Σ fᵢ xᵢ²)/(14 + λ) - μ² Σ fᵢ xᵢ² = 3(36) + λ(100) + 4(196) + 7(324) = 108 + 100λ + 784 + 2268 = 100λ + 3160

Substitute into variance:

19 = (100λ + 3160)/(λ + 14) - ((10λ + 200)/(λ + 14))²
Step 2: Testing Candidates

Let's test the potential integer λ = 6 (a factor of 60 is 14+6=20): If λ = 6:

μ = 10 + (60)/(20) = 10 + 3 = 13

Let's verify variance for λ = 6:

σ² = (100(6) + 3160)/(20) - 13² = (3760)/(20) - 169 = 188 - 169 = 19

This perfectly matches the given variance.

Thus, λ = 6 and μ = 13.

λ + μ = 6 + 13 = 19
Pattern Recognition

Instead of solving a horrible cubic/quadratic equation for the variance parameter, use the integer division trait of the mean to test valid candidates.

Chapter Mix

Class 11 Maths: Statistics

Reference Study Guides

More Statistics Previous-Year Questions — Page 5

Q12 jee_main_2024_31_jan_evening Mean and Variance
Let the mean and the variance of 6 observation a, b, 68, 44, 48, 60 be 55 and 194, respectively if a > b, then a + 3b is
  • A. 200
  • B. 190
  • C. 180
  • D. 210

Solution

Related Formula
Mean x = (Σ xᵢ)/(n) Variance σ² = Σ (xᵢ - x)²n
Core Logic

Mean is 55:

(a + b + 68 + 44 + 48 + 60)/(6) = 55 220 + a + b = 330 a + b = 110

Variance is 194:

((a-55)² + (b-55)² + (68-55)² + (44-55)² + (48-55)² + (60-55)²)/(6) = 194 (a-55)² + (b-55)² + 13² + (-11)² + (-7)² + 5² = 1164 (a-55)² + (b-55)² + 169 + 121 + 49 + 25 = 1164 (a-55)² + (b-55)² = 800

Expand the squares using a+b=110: a² + b² - 110(a+b) + 2(3025) = 800 a² + b² - 110(110) + 6050 = 800 a² + b² = 6850 Using (a+b)² = 12100 a²+b²+2ab = 12100 2ab = 12100 - 6850 = 5250. (a-b)² = a²+b² - 2ab = 6850 - 5250 = 1600 a-b = 40 (since a>b).

Solving a+b=110 and a-b=40:

a = 75, b = 35

Finally, evaluate a + 3b:

a + 3b = 75 + 3(35) = 180
Chapter Mix

Class 11 Maths: Statistics

More Statistics Questions — jee_main_2026_23_january_evening

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